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Several properties related to interior and boundary of a set
We have the following propositions, which at the first glance seem to be true, but the truth still needs some careful consideration:
 If $U$ is open, $U=\Int(\overline{U})$.
 $\Bd(\Bd(A)) = \Bd(A)$.
 $\Bd(\Int(A)) = \Bd(\overline{\Int(A)})$.
For the first proposition, $\Int(\overline{U})$ represents the union of all the open sets which are contained in $\overline{U}$. $\because U \subset \overline{U}$ and $U$ is open, $\therefore U \subset \Int(\overline{U})$. If the inverse of this proposition is true, it implies that $U$ is the maximum among the open sets that are contained in $\overline{U}$. However, this is not true in some cases. For example, let $U=(0,1) \cup (1,2)$, then $\overline{U}=[0,2]$ and $\Int(\overline{U})=(0,2) \supset U$.
For the second proposition, according to the definition of boundary, we have \begin{equation} \Bd(\Bd(A)) = \overline{\Bd(A)}  \Int(\Bd(A)) = \Bd(A)  \Int(\Bd(A)) \end{equation} If $\Bd(\Bd(A)) = \Bd(A)$, we must have $\Int(\Bd(A)) = \Phi$. It is quite tempting to say that the interior of a boundary of a set is empty. However, this is not always true. For example, let $A=(0,1) \cap \mathbb{Q}$, where $\mathbb{Q}$ is the set of rational numbers. Then \begin{equation} \Bd(A) = \overline{A} \cap \overline{XA} = [0,1] \end{equation} Of course, its interior is not an empty set. It can be imagined that the elements of the set $A$ pervade all over the space like a sponge immersed in the water. Everywhere in the space is its boundary and the interior of the space is not empty.
For the third proposition, because $\Int(A)$ is the interior of the closed set $\overline{\Int(A)}$, we have $\Int(\overline{\Int(A)})=\Int(A)$. \begin{equation}\begin{split} &\because \overline{U}=\overline{\overline{U}}=\Int(\overline{U}) \cup \Bd(\overline{U}) = U \cup \Bd(\overline{U}),\; \overline{U} = U \cup \Bd(U) &\therefore \Bd(\overline{U}) = \Bd(U) \end{split} \end{equation}
Posted in Topology
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Kuratowski's complementclosure theorem
Consider the power set $\mathcal{P}(X)$ of the topological space $X$. Given an arbitrary element in $\mathcal{P}(X)$, successive applications of the two operations: complement and closure can generate a series of subsets of $X$. The Kuratowski theorem states that the maximum number of distinct sets which can be generated from these operations is 14.
Prove: At first, we need the definitions of interior and boundary of a set to separate the whole space $X$, which can be written as: \begin{equation} X = \Int(A) \cup \Bd(A) \cup \Int(XA) \end{equation} This can be proved as follows. We've already know that \begin{equation} \overline{A} = \Bd(A)\cup\Int(A),; \overline{XA} = \Bd(XA)\cup\Int(XA) \end{equation} Then we have \begin{equation} \overline{A}\cup\overline{XA} = \left(\Int(A)\cup\Bd(A)\right) \cup \left(\Bd(XA) \cup \Int(XA)\right) \end{equation} $\because A\subset\overline{A}$ and $XA\subset\overline{XA}$, $\therefore A\cup XA = X \subset \overline{A}\cup\overline{XA}$. And $\because \overline{A}\cup\overline{XA}\subset X,, \therefore \overline{A}\cup\overline{XA}=X$. In addition, we have \begin{equation} \begin{split} \Bd(A) &= \Bd(XA) \Int(A)\cap\Bd(A) &= \Phi \Int(XA)\cap\Int(A) &= \Phi \Int(XA)\cap\Bd(XA) &= \Phi \end{split} \end{equation} Then we can separate the whole space $X$ using two interiors and one common boundary: \begin{equation} X=\Int(A)\cup\Bd(A)\cup\Int(XA)=\Int(A)\cup\Bd(XA)\cup\Int(XA) \end{equation} According to this conclusion, we can obtain the following: \begin{equation} \begin{split} X\overline{A} &= \Int(XA) X\Int{A} &= \overline{XA} \end{split} \end{equation} If we use symbol $c$ to represent the operation of complement and symbol $f$ to represent the operation of closure, the second equation in the above is actually the following: \begin{equation} \Int{A}=cfc(A) \end{equation} That is to say, after three successive operations of complement and closure, an arbitrary set $A$ can be transformed into its interior.
Next, we'll prove that $fcfcfcfc(A) = fcfc(A)$. If we use symbol $i$ to represent the operation of taking the interior of a set, we can write this identity as $fcfcfi(A)=fi(A) \Leftrightarrow fifi(A)=fi(A)$.
Another theorem is needed for proving this identity:
Theorem If $U$ is an interior of a closed set $Z$ in $X$, then $\Int(\overline{U})=U$.
This theorem can be proved like this: $\because U \subset Z$ and $Z$ is closed, $\therefore \overline{U} \subset \overline{Z} = Z$. And $\therefore \Int{\overline{U}} \subset \Int{Z} = U$. Because $\Int{\overline{U}}$ represents the union of all the open sets which are contained in $\overline{U}$ and $U$ is itself open, $\therefore U \subset \Int{\overline{U}}$. So we have $\Int{\overline{U}}=U$.
According to the above theorem, we know that $i(A)$ is the interior of the closed set $fi(A)$. Therefore, $if(i(A))=i(A)$ and $fifi(A)=fi(A)$ is proved.
Now, we know that by starting from complement operation and applying complement and closure successively to the set $A$, the maximum number of distinct sets generated in a chain (including $A$ itself) is 8. If we start from closure operation on the set $A$, at most 6 distinct sets can be generated, because as shown in the following identify \begin{equation} fcfcfcf(A)=fcfcfcfc(c(A))=fcfc(c(A)) \end{equation} after 7 times of successive operations, a duplicate set appears which is equal to $fcfc(c(A))$. Therefore, the maximum number of distinct sets in this chain is 6. The total maximum number of distinct sets is 14.
An example for the 14set in $\mathbb{R}$ is $A=(0,1) \cup (1,2) \cup {3} \cup ([4,5] \cup \mathbb{Q})$.
Posted in Topology
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